A 286 kg crate hangs from the end of a rope of length L = 13.0 m. You push horiz
ID: 1287324 • Letter: A
Question
A 286 kg crate hangs from the end of a rope of length L = 13.0 m. You push horizontally on the crate with a varying force Upper F Overscript right-arrow EndScripts to move it distance d = 4.59 m to the side (see the figure). (a) What is the magnitude of Upper F Overscript right-arrow EndScripts when the crate is in this final position? During the crate's displacement, what are (b) the total work done on it, (c) the work done by the gravitational force on the crate, and (d) the work done by the pull on the crate from the rope? (e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force does on the crate.
A 286 kg crate hangs from the end of a rope of length L = 13.0 m. You push horizontally on the crate with a varying force Upper F Overscript right-arrow EndScripts to move it distance d = 4.59 m to the side (see the figure). (a) What is the magnitude of Upper F Overscript right-arrow EndScripts when the crate is in this final position? During the crate's displacement, what are (b) the total work done on it, (c) the work done by the gravitational force on the crate, and (d) the work done by the pull on the crate from the rope? (e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force does on the crate.Explanation / Answer
From the given figure
The vertical angle is determined as follows
Sin? = d / l
or ? = Sin-1( 4.59 / 13 ) = 20.67 o
Now the tension in the string resolve into components
The vertical component supports the weight
TCos? = mg
0r T = 286*9.8 / Cos20.67 = 2998.8 N
Therefore the horizontal force F = TSin 20.67
0r F = 1058 N
b) The total work done W = 0 asthere is no change in KE
c) The work done by gravity Wg =Fs.d = - mgh
0r Wg = - mg l ( 1 - Cos? )
= -286*9.8*13 ( 1 - Cos20.67)
= - 2.347 kJ
d) asthe pull is perpendicular to the direction ofmotion, the work done = 0
e) Workdone by the Force on the crate WF = - Wg
f) Here the work done by force is not equal to F*d
and it is equal to product of the cos angle and F*d
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